User:Eml4500.f08.gravy.jad/Notes

FEA Notes and Homework 2 

Truss Problem description
Take a specific example to see how the recipe works.



Data:

Element Length

$$\ L^{(1)}=4 \,$$

$$\ L^{(2)}=2 \,$$

Young's Modules

$$\ E^{(1)}=3 \,$$

$$\ E^{(2)}=5 \,$$

Cross-sectional Area

$$\ A^{(1)}=1 \,$$

$$\ A^{(2)}=2 \,$$

Inclination angle


 * $$\ \theta^{(1)} = 30^o \,$$


 * $$\ \theta^{(2)} = -45^o \,$$

Solving a Truss Problem with Statics


First take the sum of the moments about global node 3 and solve for $$R_3$$ which is the reaction at global node 3 in the y-direction

$$ \sum {M_1}= P(4cos(30)) +R_3(4cos(30) + 2sin(45))$$

$$R_3= \frac{-P(4cos(30))} {4cos(30) + 2cos(45)} = -4.4378$$

Then use this information to take the sum of the forces in the y-direction and solve for $$R_1$$ which is the reaction at global node 1 in the y-direction.

$$\sum{F_y}= -4.4378 + P+R_1$$

$$\ R_1= -2.5622 \,$$

By observation it can be determined that there are no forces in the x-direction so $$R_1$$ and $$R_3$$ are the only reaction forces

Solving a Truss Problem with Finite Element Analysis
1)Global Picture



$$\begin{Bmatrix} f_1 \\\vdots & \\ f_6 \end{Bmatrix}_{1X6}=\begin{bmatrix} & &  \\  & k & \\ & &   \end{bmatrix}_{6X6}*    \begin{Bmatrix} d_1 \\\vdots & \\ d_6 \end{Bmatrix}_{1X6}$$

$$\begin{matrix} global& &global& &global \\ force & &stiffness& &displacement \\  column &   &matrix  & &column \\ matrix & & & &matrix \end{matrix}$$

2) Element Picture

3) Global FD at element level:

$$\ k^{(e)} d^{(e)}=f^{(e)} \,$$

$$\ k^{(e)}\,$$ = the element stiffness matrix

$$\ d^{(e)}\,$$= the element displacement matrix

$$\ f^{(e)}\,$$= the element force matrix

$$k^{(e)}\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -(l^{(e)}m^{(e)})  \\  l^{(e)}m^{(e)} & (m^{(e)})^2 & -(l^{(e)}m^{(e)}) & -(m^{(e)})^2  \\ -(l^{(e)})^2 & -(l^{(e)}m^{(e)}) & (l^{(e)})^2 & l^{(e)}m^{(e)}\\ -(l^{(e)}m^{(e)}) & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2    \end{bmatrix}_{4X4} = \underline{k^{(e)}}$$

$$\ k^{(e)}= \,$$ $$\frac{E^{(e)}A^{(e)}}{L^{(e)}}$$ axial stiffness of bar element "e" where e=1,2

$$\ l^{(e)},m^{(e)} = \,$$ director cosines of x axis (goes from 1 to 2) with respect to global (x,y coordinates)



$$\ l^{(e)}= \vec \tilde{i} \cdot \vec i= cos\theta^{(e)}\,$$

$$\ m^{(e)}= \vec i \cdot \vec j= cos\left (\frac{\pi}{2}- \theta^{(e)} \right )= sin\left (\theta^{(e)} \right )\,$$

$$\ \vec \tilde{i}=cos\left(\theta^{(e)} \right)\vec i + sin\left( \theta^{(e)}  \right )\vec j \,$$

$$\ \vec \tilde{i} \cdot \vec i = \vec \tilde{i} \left( cos\left(\theta^{(e)} \right)\vec i + sin\left( \theta^{(e)}  \right )\vec j \right) i= cos\left(\theta^{(e)}  \right) \vec i \cdot \vec i + sin\left( \theta^{(e)}  \right ) \vec i  \,$$

$$\ \vec \tilde{i} \cdot \vec j= HW \,$$

Element 1:
$$\ \theta^{(1)} = 30^o \,$$

$$\ l^{(1)}=cos\theta^{(1)}= cos 30^o= \frac{\sqrt{3}}{2} \,$$

$$\ m^{(1)}=sin\theta^{(1)}= sin 30^o= \frac{1}{2} \,$$

$$\ k^{(1)}= \frac{E^{(1)}A^1}{L^1} \frac{\left(3\right ) \left(1\right )}{4}= \frac {3}{4}\,$$

$$\ \underline{k^{(1)}}=\begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)}  \\  k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)}  \\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)} & k_{34}^{(1)}\\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)} & k_{44}^{(1)}    \end{bmatrix}_{4X4}= \left [k_{ij}^1 \right ]_{4X4} \,$$

i=row number =1,2....4

j=column number =1,2....4

$$\ k_{11}^{(1)}= k^{(1)}\left( l^{(1)} \right)^2=\frac{3}{4}*\frac{\sqrt{3}}{2}= \frac{9}{16} \,$$

$$\ k_{12}^{(1)}= k^{(1)}\left( l^{(1)}m^{(1)} \right)=\frac{3}{4}*\frac{1}{2}= \frac{3\sqrt{3}}{16} \,$$

$$\ k_{42}^{(1)}=k_{24}^{(1)}= -k^{(1)}\left( m^{(1)} \right)^2=\frac{-3}{4}*\frac{1}{2}= \frac{-3}{16} \,$$

Observation:

1) Only need to compute 3 numbers. Other coefficients have same absolute value, just differ by (+ -)

2) Matrix $$ \underline{k^{(1)}}$$ is synmetrix, $$k_{ij}^{(1)}=k_{ji}^{(1)}$$

$$k_{13}^{(e)}=k_{31}^{(e)}$$ (just interchange the row and column index)

$$\ \underline{k^{(1)}}=\begin{bmatrix} \frac{3}{4}(\frac{\sqrt{3}}{2})^2 & \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & -\frac{3}{4}(\frac{\sqrt{3}}{2})^2 & -\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2})  \\  \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & \frac{3}{4}(\frac{1}{2})^2 & -\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & -\frac{3}{4}(\frac{1}{2})^2  \\ -\frac{3}{4}(\frac{\sqrt{3}}{2})^2 & -\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & \frac{3}{4}(\frac{\sqrt{3}}{2})^2 & \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2})\\-\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & -\frac{3}{4}(\frac{1}{2})^2 & \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & \frac{3}{4}(\frac{1}{2})^2    \end{bmatrix}_{4X4}=  \underline{k^{(1)}} _{4X4} \,$$

$$\ \underline{k^{(1)}}=\begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}  \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}  \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16}    \end{bmatrix}_{4X4}=  \underline{k^{(1)}} _{4X4} \,$$

In general, $$k_{ij}^{(e)}=k_{ji}^{(e)} or k^{(e) T} =k^{(e)}$$

The transpose of $$ k^{(e)}$$ is equal to $$ k^{(e)}$$

$$\ \underline{k^{1}}=\begin{bmatrix} k_{11}^{(e)} & k_{12}^{(e)} & k_{13}^{(e)} & k_{14}^{(e)}  \\   & k_{22}^{(e)} & k_{23}^{(e)} & k_{24}^{(e)}  \\  &  & k_{33}^{(e)} & k_{34}^{(e)}\\ sym. & &  & k_{44}^{(e)}    \end{bmatrix}_{4X4}= \left [k_{ij}^1 \right ]_{4X4} \,$$

Element 2:
$$\ k^{(2)}= \frac{E^{(2)}A^{(2)}}{L^{(2)}} \frac{\left(5\right ) \left(2\right )}{2}= 5\,$$

$$\ \theta^{(2)} = \frac{-\pi}{4}= -45^o \,$$

$$\ l^{(2)}=cos\theta^{(2)}= cos (\frac{-\pi}{4})= \frac{\sqrt{2}}{2} \,$$

$$\ m^{(2)}=sin\theta^{(2)}= sin (\frac{-\pi}{4})= \frac{-\sqrt{2}}{2} \,$$

$$\ k^{(2)} = \left [k_{ij}^{(2)} \right ]_{4X4} \,$$

$$\ k_{11}^{(2)} = k^{(2)}(l^{(2)})^2 = 5\frac{2}{4} = 2.5 \,$$

Observation:

1) Absolute value of all coefficients $$\ k_{ij}^{(e)} \,$$, $$\ e=2 \,$$ , $$\ (i,j)=1, ..., 4 \,$$ are the same => comp 1 coefficient. For other coefficients, and (+) or (-)

2) $$\ k ^{(2)T} = k^{(2)} \,$$ ie. $$\ k^{(2)} \,$$ sym

$$\ \underline{k^{(2)}}=\begin{bmatrix} \frac{5}{2}& -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}  \\  -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}  \\ -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}\\ \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}    \end{bmatrix}_{4X4}=  \underline{k^{(2)}} _{4X4} \,$$

Elem FD rel: $$\ k^{(e)}d^{(e)}= f^{(e)} \,$$
 * e=1,2

$$\begin{Bmatrix} d_1 \\\vdots & \\ d_6 \end{Bmatrix}_{1X6}, \begin{Bmatrix} f_1 \\\vdots & \\ f_6 \end{Bmatrix}_{1X6} $$

Global FD relationship:

$$\ \underline{K}_{nxn} \underline{d}_{nx1} = \underline{F}_{nx1} \,$$

there, n=6

$$\ \begin{bmatrix} k_{11} & k_{12} & k_{13} & k_{14} & k_{15} & k_{16}  \\ k_{21}& k_{22} & k_{23} & k_{24} & k_{25} & k_{26} \\ k_{31}& k_{32} & k_{33} & k_{34} & k_{35} & k_{36} \\ k_{41}& k_{42} & k_{43} & k_{44} & k_{45} & k_{46} \\ k_{51}& k_{52} & k_{53} & k_{54} & k_{55} & k_{56} \\ k_{61}& k_{62} & k_{63} & k_{64} & k_{65} & k_{66}  \end{bmatrix}  \begin{Bmatrix}  d_{1} \\ d_{2} \\ d_{3} \\ d_{4} \\ d_{5} \\ d_{6} \end{Bmatrix} = \begin{Bmatrix}  F_{1} \\ F_{2} \\ F_{3} \\ F_{4} \\ F_{5} \\ F_{6} \end{Bmatrix}\,$$

Global F-D Relationship
Noting the fact that the equations for the elemental $$\underline{d^{(e)}}$$ and global $$\underline{d}$$ overlap, as seen in the below picture, the following equations can be derived.



$$d_1 = d_1^{(1)}$$ $$d_2 = d_2^{(1)}$$ $$d_3 = d_3^{(1)} = d_1^{(2)}$$ $$d_4 = d_4^{(1)} = d_2^{(2)}$$ $$d_5 = d_3^{(2)}$$ $$d_6 = d_4^{(2)}$$

Below is a representation of the global $$\underline{K}$$ matrix, which is the stiffness matrix for the system. The two colored boxes represent the contributions of the elemental $$\underline{k^{(e)}}$$. The $$\underline{k^{(1)}}$$ matrix is shown in blue, while the red box represents $$\underline{k^{(2)}}$$.

$$\ \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}\,$$

EX:

$$\ K_{11} = k_{11}^{(1)} \,$$

$$\ K_{33} = k_{33}^{(1)} + k_{11}^{(2)} \,$$

$$\ K_{43} = K_{34} = k_{34}^{(1)} + k_{12}^{(2)} = k_{43}^{(1)} + k_{21}^{(2)} \,$$

$$\ k_{11}^{(1)} = \frac{9}{16} \,$$

$$\ k_{12}^{(1)} = \frac{3\sqrt{3}}{16}$$

$$ K_{33} = \frac{9}{16} + \frac{5}{2} = 3.0625 $$

$$ K_{43} = \frac{3\sqrt{3}}{16} + \frac{-5}{2} = 2.1752 $$

$$ K_{44} = k_{44}^{(1)} + k_{22}^{(2)} = \frac{3}{16} + \frac{5}{2} = 2.6875 $$ Mbr> $$\ \ = \begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & (\frac{9}{16} + \frac{5}{2}) & (\frac{3\sqrt{3}}{16} - \frac{5}{2}) & -\frac{5}{2} & \frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & (\frac{3\sqrt{3}}{16} - \frac{5}{2}) & (\frac{3}{16} + \frac{5}{2}) & \frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2} \\ 0 & 0 & \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$ $$\ \ = \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3428 & 0 & 0 \\  0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0 \\ -0.5625 & -0.3248 & (0.5625 + 2.5) & (0.3248 - 2.5) & -2.5 & 2.5 \\ -0.3248 & -0.1875 & (0.3248 - 2.5) & (0.1875 + 2.5) & 2.5 & -2.5 \\  0 & 0 & -2.5 & 2.5 & 2.5 & -2.5 \\ 0 & 0 & 2.5 & -2.5 & -2.5 & 2.5  \end{bmatrix}_{6X6}\,$$

Elimination of Known Degrees of Freedom
Knowing that global nodes 1 and 3 are actually fixed points, it is seen that there is no possibility of them moving. Therefore, the displacement at those points is zero.

$$\ d_1 = d_2 = d_5 = d_5 = d_6 = 0 \,$$

$$\ \underline{d}=\begin{Bmatrix} d_1 \\ d_2  \\  d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix}= \begin{Bmatrix}  0 \\ 0 \\  d_3 \\ d_4 \\ 0 \\ 0 \end{Bmatrix}$$

Since 4 of the values for $$\underline{d}$$ are zero, we can reduce the system. Turning $$\underline{d}$$ into a 2x2 matrix, would mean that you would be able to reduce $$\underline{K}$$ to a 6x2 matrix. This is down by removing the columns that are associated with the rows within $$\underline{d}$$ that are zero. The new $$\underline{K}$$ is shown below.

$$\ \underline{K}=\begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24} \\  K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{bmatrix}$$

This makes the F-D equation: $$\ \underline{K}_{6x2} \underline{d}_{2x1} = \underline{F}_{6x1} \,$$

The next step is to see that since global nodes 1 and 3 are fixed, we can able the concept of "Virtual Work" which will allow us to ignore their effects for the moment and let us turn $$\underline{F}$$ into a 2x1 by removing the components from those fixed nodes. However, by doing this, we need to adjust $$\underline{K}$$ to keep the equation true. We can further reduce $$\underline{K}$$ down to a 2x2, and the following equation results:

$$\ \underline{K}_{2x2} * \underline{d}_{2x1} = \underline{F}_{2x1} \,$$

Where $$\underline{K}$$ is equal to:

$$\ \underline{K}=\begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \end{bmatrix}$$ Since $$\underline{F}$$ is now only made up of the forces from global node 2, we now know the value of $$\underline{F}$$. $$\ \underline{F} = \begin{Bmatrix} F_3 \\ F_4 \end{Bmatrix}= \begin{Bmatrix}  0 \\ P \end{Bmatrix}$$ We also can solve for $$\underline{K}$$: $$\ \underline{K}=\begin{bmatrix} 3.0625 & 2.1752 \\ 2.1752 & 2.6875 \end{bmatrix}$$

At this point, the only term not known in the global F-D equation is $$\underline{d}$$. In order to solve the system, we must first solve for $$\underline{d}$$, which gives the following equation: $$\ \underline{d}_{2x1} = \underline{K}^{-1}_{2x2} * \underline{F}_{2x1} \,$$

Where $$\underline{K^{-1}}$$ is equal to: $$\ \underline{K^{-1}} = \frac{1}{det(\underline{K})}*\begin{bmatrix} K_{44} & -K_{43} \\ -K_{34} & K_{33} \end{bmatrix}$$ $$\ det(\underline{K}) = K_{33} * K_{34} - K_{34} * K_{43} $$

$$\ \underline{K^{-1}} = \begin{bmatrix} 0.7681 & -0.6217 \\ -0.6217 & 0.8753 \end{bmatrix}$$ Plugging everything in and solving for $$\underline{d}$$ gives:

$$\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix} = \underline{K^{-1}} \begin{Bmatrix} 0 \\ P \end{Bmatrix}$$

$$\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix} = \begin{bmatrix} 0.7681 & -0.6217 \\ -0.6217 & 0.8753 \end{bmatrix} * \begin{Bmatrix} 0 \\ P \end{Bmatrix} = \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix} $$

$$\ \underline{d} = \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

Computing Reactions
$$ Elem 1: \underline{k}^{(1)} \underline{d}^{(1)} = \underline{f}^{(1)} $$

$$\ d^{(1)} = \begin{bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{bmatrix}_{4X1}\,$$

$$\ = \begin{bmatrix} 0.5625 & 0.32476 & ... & ... \\ & & & \\ & & & \\ & & & \end{bmatrix}_{4X4} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}_{4X1} \,$$

$$\ = \begin{bmatrix} -0.5625 & -0.32476 \\ ... & ... \\ & \\ & \end{bmatrix}_{4X2} \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}_{4X1} \,$$

$$\ = \begin{bmatrix} -4.4378 \\ -2.5622 \\ +4.4378 \\ +2.5622 \end{bmatrix}_{4X1}\,$$

$$\ = \begin{bmatrix} f_{1}^{(1)} \\ f_{2}^{(1)} \\ f_{3}^{(1)} \\ f_{4}^{(1)} \end{bmatrix}_{4X1}\,$$


 * f1(1) and f2(1) = Reactions


 * f3(1) and f4(1) = Internal Forces


 * Observation: Element 1 is in equilibrium


 * ΣFx = f1(1) + f3(1)


 * ΣFy = f2(1) and f4(1)



ΣMat any point = 0

$$P_{1}^{(1)} = [(f_{1}^{(1)})^2 + (f_{2}^{(1)})^2]^ \frac{1}{2}$$

$$ Elem 2: \underline{k}^{(2)} \underline{d}^{(2)} = \underline{f}^{(2)} $$


 * The same method to solve for element one can also be used to solve for element 2

HW: Verify equilibrium of element 1
Let $$P_{1}^{(1)} = f_{1}^{(1)} + f_{2}^{(1)} = -4.4378 i\hat{} -2.5622j\hat{}$$ and let $$P_{2}^{(1)} = f_{3}^{(1)} + f_{4}^{(1)} = 4.4378 i\hat{} + 2.5622j\hat{}$$

Note that $$P_{1}^{(1)} + P_{2}^{(1)} = 0$$ which illustrates that $$\sum{}F=0$$.

Also, note that $$P_{1}^{(1)} \times P_{2}^{(1)} = \left( -4.4378 i\hat{} -2.5622j\hat{} \right) \times \left( 4.4378 i\hat{} +2.5622j\hat{} \right) = \left( \left( -4.4378 \times 2.5622 \right) - \left( 4.4378 \times -2.5622 \right) \right) k\hat{} = 0$$.

This shows that $$P_{1}^{(1)}$$ and $$P_{2}^{(1)}$$ are in line with each other. Since all forces are in line with each other, $$\sum{}M_{anypoint} = 0$$.

HW: Verify equilibrium of node 2
Note that $$\sum{}F = P_{1}^{(1)} + P_{2}^{(2)} + P = \left( -4.4378 i\hat{} -2.5622j\hat{} \right) + \left( 4.4378 i\hat{} -4.4378j\hat{} \right) + 7j\hat{} = 0$$.

Also, since all forces are acting on the same point, $$\sum{}M_{2} = 0$$.

Contributing Team Members
The following students contributed to this report:

Jessica Dyer Eml4500.f08.gravy.jad 22:11, 25 September 2008 (UTC)

Mikal McFarlane Eml4500.f08.gravy.mmm 00:52, 26 September 2008 (UTC)

Patrick Clayton Eml4500.f08.gravy.psc 16:33, 26 September 2008 (UTC)

Brian Eggeman Eml4500.f08.gravy.bje 17:32, 26 September 2008 (UTC)

Justin Tennant Eml4500.f08.gravy.JMT 16:31, 26 September 2008 (UTC)

Sean Sullivan Eml4500.f08.gravy.sms 17:20, 26 September 2008 (UTC)